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25 Mar 2019, 00:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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73% (01:56) correct
27%
(02:19)
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based on 182
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A subset B of the set of integers from 1 to 100, inclusive, has the property that no two elements of B sum to 125. What is the maximum possible number of elements in B?
(A) 50
(B) 51
(C) 62
(D) 65
(E) 68
(A) 50
(B) 51
(C) 62
(D) 65
(E) 68
25 Mar 2019, 00:25
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Bunuel
Primary Set of Integers = {1, 2, 3, 4, 5.....100}
The Unwanted pairs making sum 125 = {25, 26, 27, 28, 29,..........96, 97, 98, 99, 100}
Now, The set making pairs making sum 125 has 100-24 = 76 terms
But we can use half of the terms in the required set and leave the other half
So the requires set will have all terms from 1 to 24 and half of 76 terms i.e. 38 terms
Total Integers in Set B = 24+38 = 62
Answer: Option C
Set A = { 1,2,3,4,5...,99,100}
Set B = given when 2 numbers added it will be less than 125
If I divide 125/2 = 62.5 , (If I add 62+63 = 125 )
So we can not consider 63 in the Subset , this gives us the total numbers we can consider are till 62
set B = { 1,2,3,4,5,6,7....60,61,62}.
Set B = given when 2 numbers added it will be less than 125
If I divide 125/2 = 62.5 , (If I add 62+63 = 125 )
So we can not consider 63 in the Subset , this gives us the total numbers we can consider are till 62
set B = { 1,2,3,4,5,6,7....60,61,62}.
